I have the matrix $$ M= \begin{pmatrix} a^2&a\vec{v}^T\\ a\vec{v}&B+\vec{v}\vec{v}^T \end{pmatrix}, $$ where $a\in\mathbb{R}$, $\vec{v}\in\mathbb{R}^n$, and $B\in\mathbb{R}^{n\times n}$.
Since $$ M^{-1}=\frac{1}{a^2} \begin{pmatrix} 1+\vec{v}^TB^{-1}\vec{v}&-a\vec{v}^TB^{-1}\\ -aB^{-1}\vec{v}&a^2B^{-1} \end{pmatrix}, $$ using the orthogonal projection $$ P= \begin{pmatrix} 0&\vec{0}^T\\ \vec{0}&I \end{pmatrix}, $$ with $I$ the $n\times n$ identity matrix, it is easy to show that $M^{-1}$ and $B^{-1}$ have their eigenvalues interlaced, i.e., $m_1^{-1}\geq b_1^{-1}\geq m_2^{-1}\geq b_2^{-1}\geq\cdots m_{n}^{-1}\geq b_{n}^{-1}\geq m_{n+1}^{-1}$, where $m_i$ is the $i$th eigenvalue of $M$, and $b_i$ that of $B$. Therefore, $m_1\leq b_1\leq m_2\leq b_2\leq\cdots m_{n}\leq b_{n}\leq m_{n+1}$, and it can be stated that the eigenvalues of $M$ and $B$ are also interlaced.
Is it possible to arrive at the same conclusion without resorting to inverting matrices? In other words, what orthogonal projection acting upon $M$ will produce $B$?
Here's another approach to showing that the eigenvalues interlace. Instead of using Cauchy's interlacing theorem, use Weyl's inequalities. Note that we can write $M = B_0 + R$, where $B_0$ and $R$ are given by $$ B_0 = \pmatrix{0\\&B}, \quad R = \pmatrix{a^2 & av^T\\av & vv^T} = \pmatrix{a\\ v} \pmatrix{a\\v}^T. $$ If $m_1\leq \cdots \leq m_{n+1}$ are the eigenvalues of $M$ and $\beta_1 \leq \cdots \leq \beta_{n+1}$ are the eigenvalues of $B_0$, then because $R$ has one positive eigenvalue and $n$ zero eigenvalues, we have $\beta_i \leq m_i \leq \beta_{i+1}$.
On the other hand, the eigenvalues of $B_0$ are simply the eigenvalues of $B$ with an extra $0$ eigenvalue.